How do you solve #Log_(5)x + log_(3)x = 1#?

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Answer 1 · 2016-06-27T19:07:39

#x = 3^{(log_e 5)/(log_e 15)}#

We know that

#log_a b = (log_eb)/(log_e a)#

Then

#Log_(5)x + log_(3)x = 1->(log_e x)/(log_e 5)+(log_e x)/(log_e 3)=1#

or

#log_e x= (log_e 5 cdot log_e 3)/(log_e 5 + log_e 3)#

or

#x = e^{(log_e 5 cdot log_e 3)/(log_e 5 + log_e 3)}#

#x = 3^{(log_e 5)/(log_e 15)}#